The electric field associated with a monochromatic light wave is given by $E = E_0 \sin \left[ \left( 1.57 \times 10^7 \text{ m}^{-1} \right) (x - ct) \right]$. The stopping potential when this light is used in a photoelectric experiment with a metal having a work function of $1.9 \text{ eV}$ is: (Planck's constant,$h = 6.64 \times 10^{-34} \text{ J-s}$) (in $\text{ V}$)

The maximum velocity of emitted photoelectrons is $4.8 \ m/s$. If the $e/m$ ratio of an electron is $1.76 \times 10^{11} \ C/kg$,then the stopping potential is ......

$A$ metal surface is illuminated by light of a given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become

In the photoelectric effect, an electromagnetic wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is $2.14 \text{ eV}$ and the stopping potential is $2 \text{ V}$, what is the wavelength of the electromagnetic wave (in $\text{ nm}$)? (Given $hc = 1242 \text{ eV nm}$, where $h$ is Planck's constant and $c$ is the speed of light in vacuum.)

For zero photoelectric current,the stopping potential is:

Radiation of two photon energies,twice and five times the work function of a metal,are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases is: